Formal geometry proofs

Start with the fact that the interior angles of a triangle sum to $180°$.

We need to find two different unknown angles in the diagram. We can find them using facts we know about triangles.

One of the unknown angles, $x$, is inside of the triangle. So we can use the fact that the interior angles of a triangle have a sum of $180°$. Write an equation for this and then solve for $x$.

Angle $x$ has a measure of $\text{87,27}°$.

The other angle we need to find, $y$, is outside of the triangle. It is an exterior angle because it is made by the extension of one of the sides of the triangle. To solve for $y$ use the theorem for the exterior angles of a triangle.

The exterior angle theorem for triangles tells us that an exterior angle is equal to the sum of the two interior angles opposite the exterior angle. In this triangle, the exterior angle $Q\hat{M}P$ is equal to the sum of the opposite interior angles, $\angle N$ and $\angle P$. The figure below shows this with shaded angles. If we add the orange and blue angles inside the triangle together, we will get an angle which is exactly the same size as the exterior angle.

We can write an equation based on the exterior angle theorem to find the value of $y$:

Exterior angle of a triangle = The sum of the opposite interior angles

The correct answers are:

1. $x =$ $\text{87,27}$${°}$
2. $y =$ $\text{129,02}$${°}$

Start with the fact that the interior angles of a triangle sum to $180°$.

We need to find two different unknown angles in the diagram. We can find them using facts we know about triangles.

One of the unknown angles, $x$, is inside of the triangle. So we can use the fact that the interior angles of a triangle have a sum of $180°$. Write an equation for this and then solve for $x$.

Angle $x$ has a measure of $\text{38,94}°$.

The other angle we need to find, $y$, is outside of the triangle. It is an exterior angle because it is made by the extension of one of the sides of the triangle. To solve for $y$ use the theorem for the exterior angles of a triangle.

The exterior angle theorem for triangles tells us that an exterior angle is equal to the sum of the two interior angles opposite the exterior angle. In this triangle, the exterior angle $D\hat{A}C$ is equal to the sum of the opposite interior angles, $\angle B$ and $\angle C$. The figure below shows this with shaded angles. If we add the orange and blue angles inside the triangle together, we will get an angle which is exactly the same size as the exterior angle.

We can write an equation based on the exterior angle theorem to find the value of $y$:

Exterior angle of a triangle = The sum of the opposite interior angles

The correct answers are:

1. $x =$ $\text{38,94}$${°}$
2. $y =$ $\text{77,88}$${°}$

Start with the fact that the interior angles of a triangle sum to $180°$.

We need to find two different unknown angles in the diagram. We can find them using facts we know about triangles.

One of the unknown angles, $x$, is inside of the triangle. So we can use the fact that the interior angles of a triangle have a sum of $180°$. Write an equation for this and then solve for $x$.

Angle $x$ has a measure of $\text{100,29}°$.

The other angle we need to find, $y$, is outside of the triangle. It is an exterior angle because it is made by the extension of one of the sides of the triangle. To solve for $y$ use the theorem for the exterior angles of a triangle.

The exterior angle theorem for triangles tells us that an exterior angle is equal to the sum of the two interior angles opposite the exterior angle. In this triangle, the exterior angle $Q\hat{M}P$ is equal to the sum of the opposite interior angles, $\angle N$ and $\angle P$. The figure below shows this with shaded angles. If we add the orange and blue angles inside the triangle together, we will get an angle which is exactly the same size as the exterior angle.

We can write an equation based on the exterior angle theorem to find the value of $y$:

Exterior angle of a triangle = The sum of the opposite interior angles

The correct answers are:

1. $x =$ $\text{100,29}$${°}$
2. $y =$ $\text{79,71}$${°}$

The diagram below represents a polygon with 6 sides. This polygon is called a hexagon.

The sum of the interior angles of this hexagon = 720 ${°}$.

The diagram below represents a polygon with 3 sides. This polygon is called a triangle.

The sum of the interior angles of this triangle = 180 ${°}$.

The diagram below represents a polygon with 7 sides. This polygon is called a heptagon.

The sum of the interior angles of this heptagon = 900 ${°}$.

The polygon shown in the question has $\text{8}$ sides. This is called a ocatagon.

We are told that this is a regular polygon. That means that all of the sides have the same length, and all of the angles have the same size.

So we can use the following formula to calculate the size of the exterior angle:

where $n$ is the number of sides. So

The polygon shown in the question has $\text{3}$ sides. This is called a triangle.

We are told that this is a regular polygon. That means that all of the sides have the same length, and all of the angles have the same size.

So we can use the following formula to calculate the size of the exterior angle:

where $n$ is the number of sides. So

The polygon shown in the question has $\text{4}$ sides. This is called a quadrilateral.

We are told that this is a regular polygon. That means that all of the sides have the same length, and all of the angles have the same size.

So we can use the following formula to calculate the size of the exterior angle:

where $n$ is the number of sides. So

A plane 2D shape with straight lines that enclose, (one side follows the other side until it forms a closed shape) is a polygon. When all the sides and angles are the same size, it's called a regular polygon. The diagram below represents a polygon with $\text{4}$ sides. This polygon is called a quadrilateral.

Let $n$ be the number of sides. Then $n = \text{4}$.

We could also calculate the interior angle and then use supplementary rule to find $x$.

Therefore the sum of the angles in this polygon is:

We can use this to work out what each individual interior angle is.

Finally, we know that the sum of angles on a straight line is $\text{180}\text{°}$. So

So $x = \text{90}\text{°}$

A plane 2D shape with straight lines that enclose, (one side follows the other side until it forms a closed shape) is a polygon. When all the sides and angles are the same size, it's called a regular polygon. The diagram below represents a polygon with $\text{4}$ sides. This polygon is called a quadrilateral.

Let $n$ be the number of sides. Then $n = \text{4}$.

We could also calculate the interior angle and then use supplementary rule to find $x$.

Therefore the sum of the angles in this polygon is:

We can use this to work out what each individual interior angle is.

Finally, we know that the sum of angles on a straight line is $\text{180}\text{°}$. So

So $x = \text{90}\text{°}$

A plane 2D shape with straight lines that enclose, (one side follows the other side until it forms a closed shape) is a polygon. When all the sides and angles are the same size, it's called a regular polygon. The diagram below represents a polygon with $\text{4}$ sides. This polygon is called a quadrilateral.

Let $n$ be the number of sides. Then $n = \text{4}$.

We could also calculate the interior angle and then use supplementary rule to find $x$.

Therefore the sum of the angles in this polygon is:

We can use this to work out what each individual interior angle is.

Finally, we know that the sum of angles on a straight line is $\text{180}\text{°}$. So

So $x = \text{90}\text{°}$

The two triangles share a side. Which of the four congruency cases could you prove using this common side and the other given information?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $SP \; \bot \; RQ$, so we know that $S\hat{P}R = S\hat{P}Q = 90°$. So we have a pair of $\color{orange}{90°}$ angles.

We were also given that $SR = SQ$, so we have a pair of equal sides. However, these are also the hypotenuses of the triangles, so we have a pair of equal hypotenuses.

We can see that the line $SP$ is a side of both triangles. When the same side is shared by the two triangles, we say that the side is common. So, we have a pair of equal sides.

This matches the $\color{orange}{90°}$HS ($90°$, hypotenuse, side) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the same format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles.
• Conclude, and give a reason for the congruency. Make sure that you label the triangles in the matching order.

In $\color{red}{\Delta SRP}$ and $\color{red}{\Delta SQP}$:

1. $\color{gray}{SR = SQ}$ $\color{gray}{\left(\text{given}\right)}$
2. $\color{gray}{SP \text{ is common}}$
3. $\color{gray}{S\hat{P}R = S\hat{P}Q = 90°}$ $\color{gray}{\left(\text{given}\right)}$

$\color{ForestGreen}{\therefore \Delta SRP \equiv \Delta SQP}$ $\color{ForestGreen}{\left(90°\text{HS}\right)}$.

The two triangles share a side. Which of the four congruency cases could you prove using this common side and the other given information?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $SQ = PR = \text{5} \ \text{mm}$, so we have a pair of equal sides.

We can see that the line $QR$ is a side of both triangles. When a matching side is shared by two triangles, we say that the side is common. So, we have another pair of equal sides.

We were also told that $SQ \parallel RP$, so we can use alternate angles between parallel lines to prove that $S\hat{Q}R = P\hat{R}Q$. So we have a pair of equal angles.

The angle is between the two sides, so this matches the SAS (side, angle, side) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the same format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles.
• Conclude, and give a reason for the congruency. Make sure that you label the triangles in the matching order.

In $\color{red}{\Delta SQR}$ and $\color{red}{\Delta PRQ}$:

1. $\color{gray}{SQ = PR = \text{5}\ \text{mm}}$ $\color{gray}{\left(\text{given}\right)}$
2. $\color{gray}{QR \text{ is common}}$
3. $\color{gray}{S\hat{Q}R = P\hat{R}Q}$ $\color{gray}{\left(\text{alt } \angle \text{s; } SQ \parallel RP\right)}$

$\color{ForestGreen}{\therefore \Delta SQR \equiv \Delta PRQ}$ $\color{ForestGreen}{\left(\text{SAS}\right)}$.

The two triangles share a side. Which of the four congruency cases could you prove using this common side and the other given information?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $SR = QP = \text{7} \ \text{cm}$, so we have a pair of equal sides.

We can see that the line $RP$ is a side of both triangles. When a matching side is shared by two triangles, we say that the side is common. So, we have another pair of equal sides.

We were also told that $SR \parallel PQ$, so we can use alternate angles between parallel lines to prove that $S\hat{R}P = Q\hat{P}R$. So we have a pair of equal angles.

The angle is between the two sides, so this matches the SAS (side, angle, side) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the same format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles.
• Conclude, and give a reason for the congruency. Make sure that you label the triangles in the matching order.

In $\color{red}{\Delta SRP}$ and $\color{red}{\Delta QPR}$:

1. $\color{gray}{SR = QP = \text{7}\ \text{cm}}$ $\color{gray}{\left(\text{given}\right)}$
2. $\color{gray}{RP \text{ is common}}$
3. $\color{gray}{S\hat{R}P = Q\hat{P}R}$ $\color{gray}{\left(\text{alt } \angle \text{s; } SR \parallel PQ\right)}$

$\color{ForestGreen}{\therefore \Delta SRP \equiv \Delta QPR}$ $\color{ForestGreen}{\left(\text{SAS}\right)}$.

The triangles are congruent. That means they are identical. Start by using this fact to label all the sides which are not already labelled.

The first question asks us to determine the length of the side of the diagram labelled $x$. The $x$ is attached to side $WX$.

Based on the labels in the diagram, we do not know enough to calculate $x$ immediately. However, the question states that the two triangles are congruent. That means they are identical. We can use that information to label all the sides which are not already labelled. The question also states that $UW \perp XV$, which means that there are two right angles. We should label those right angles too.

Now we can see two right-angled triangles. And we can see which sides are equal to each other. Both of the triangles include a side labelled $x$.

Now calculate $x$ using the theorem of Pythagoras. We will focus on triangle $UXW$, but you can use either one, because they are congruent.

The length of $WX$ is $\text{15}\ \text{units}$.

For the second question, we need to find the length of segment $XV$. This segment is made of $XU$ and $UV$. So we can add them together to get the total length: $XV = 9 + 9 = \text{18}\ \text{units}$.

The correct answers are:

1. The length of side $x$ is $\text{15}\ \text{units}$.
2. The length of $XV$ is $\text{18}\ \text{units}$.

The triangles are congruent. That means they are identical. Start by using this fact to label all the sides which are not already labelled.

The first question asks us to determine the length of the side of the diagram labelled $x$. The $x$ is attached to side $UX$.

Based on the labels in the diagram, we do not know enough to calculate $x$ immediately. However, the question states that the two triangles are congruent. That means they are identical. We can use that information to label all the sides which are not already labelled. The question also states that $UW \perp XV$, which means that there are two right angles. We should label those right angles too.

Now we can see two right-angled triangles. And we can see which sides are equal to each other. Both of the triangles include a side labelled $x$.

Now calculate $x$ using the theorem of Pythagoras. We will focus on triangle $UXW$, but you can use either one, because they are congruent.

The length of $UX$ is $\text{9}\ \text{units}$.

For the second question, we need to find the length of segment $XV$. This segment is twice as long as $x$, which we calculated above. So $XV = 2 \left( 9 \right) = \text{18}\ \text{units}$.

The correct answers are:

1. The length of side $x$ is $\text{9}\ \text{units}$.
2. The length of $XV$ is $\text{18}\ \text{units}$.

The triangles are congruent. That means they are identical. Start by using this fact to label all the sides which are not already labelled.

The first question asks us to determine the length of the side of the diagram labelled $x$. The $x$ is attached to side $UX$.

Based on the labels in the diagram, we do not know enough to calculate $x$ immediately. However, the question states that the two triangles are congruent. That means they are identical. We can use that information to label all the sides which are not already labelled. The question also states that $UW \perp XV$, which means that there are two right angles. We should label those right angles too.

Now we can see two right-angled triangles. And we can see which sides are equal to each other. Both of the triangles include a side labelled $x$.

Now calculate $x$ using the theorem of Pythagoras. We will focus on triangle $UXW$, but you can use either one, because they are congruent.

The length of $UX$ is $\text{9}\ \text{units}$.

For the second question, we need to find the length of segment $XV$. This segment is twice as long as $x$, which we calculated above. So $XV = 2 \left( 9 \right) = \text{18}\ \text{units}$.

The correct answers are:

1. The length of side $x$ is $\text{9}\ \text{units}$.
2. The length of $XV$ is $\text{18}\ \text{units}$.

What must you do before assuming that two triangles are congruent?

We cannot use congruency to deduce that $LM = KJ$ unless we are certain that the triangles are congruent. Sometimes, we are told in the question that the triangles are congruent. However, this is not the case here. We must first prove that $\Delta LMK \equiv \Delta KJL$ before using this fact to answer the question.

Here is the completed proof:

In $\Delta LMK$ and $\Delta KJL$:

1. $\hat{J} = 89°$ $\left(\text{sum of } \angle \text{s in } \Delta\right)$
   $\quad \therefore \hat{J} = \hat{M}$
2. $M\hat{L}K = J\hat{K}L = 50°$ $\left(\text{given}\right)$
3. $LK \text{ is common}$

$\therefore \Delta LMK \equiv \Delta KJL$ $\left(\text{SAA}\right)$

$\therefore LM = KJ$ $\left(\Delta LMK \equiv \Delta KJL\right)$

Read through the proof carefully, following every step. Is anything incorrect? Is there anything missing?

This option does not have a reason for the deduction on the last line. We know that the sides are equal only because $\Delta MJK \equiv \Delta KLM$. So, we must include this on the last line.

What must you do before assuming that two triangles are congruent?

We cannot use congruency to deduce that $PS = RQ$ unless we are certain that the triangles are congruent. Sometimes, we are told in the question that the triangles are congruent. However, this is not the case here. We must first prove that $\Delta PSR \equiv \Delta RQP$ before using this fact to answer the question.

Here is the completed proof:

In $\Delta PSR$ and $\Delta RQP$:

1. $PR \text{ is common}$
2. $\hat{Q} = 79°$ $\left(\text{sum of } \angle \text{s in } \Delta\right)$
   $\quad \therefore \hat{Q} = \hat{S}$
3. $S\hat{P}R = Q\hat{R}P = 58°$ $\left(\text{given}\right)$

$\therefore \Delta PSR \equiv \Delta RQP$ $\left(\text{SAA}\right)$

$\therefore PS = RQ$ $\left(\Delta PSR \equiv \Delta RQP\right)$

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. Which of these cases could you prove from the information given?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were given that $JL = KL$, so we have a pair of equal sides.

We were given that $J\hat{L}N = 90°$. We can use vertically opposite angles to prove that $J\hat{L}N = K\hat{L}M = 90°$. So we have a pair of ${\color{MediumVioletRed}{90°}}$ angles.

We were given that $JN = KM$, so we have another pair of equal sides. However, these are also the hypotenuses of the triangles, so we have a pair of equal hypotenuses.

This matches the ${\color{MediumVioletRed}{90°}}$HS ($90°$, hypotenuse, side) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the following format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles. The working to prove each of these three sides or angles is labelled 1, 2, 3.
• Conclude, and give a reason for the congruency.

In $\color{red}{\Delta JLN}$ and $\color{red}{\Delta KLM}$:

1. $\color{Gray}{JL = KL}$ $\color{Gray}{\left(\text{given}\right)}$
2. ${\color{Gray}{\angle JLN = \angle KLM = 90°}}$ ${\color{Gray}{\left(\text{vert opp } \angle \text{s equal}\right)}}$
3. ${\color{Gray}{JN = KM}}$ ${\color{Gray}{\left(\text{given}\right)}}$

$\color{ForestGreen}{\therefore \Delta JLN \equiv \Delta KLM}$ $\color{ForestGreen}{\left(90°\text{HS}\right)}$.

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. Which of these cases could you prove from the information given?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were given that $TQ = SQ$, so we have a pair of equal sides.

We were given that $QR = QP$, so we have another pair of equal sides.

We can use vertically opposite angles to prove that $T\hat{Q}R = S\hat{Q}P$. So we have a pair of equal angles.

The angle is between the two sides, so this matches the SAS (side, angle, side) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the following format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles. The working to prove each of these three sides or angles is labelled 1, 2, 3.
• Conclude, and give a reason for the congruency.

In $\color{red}{\Delta TQR}$ and $\color{red}{\Delta SQP}$:

1. $\color{Gray}{TQ = SQ}$ $\color{Gray}{\left(\text{given}\right)}$
2. ${\color{Gray}{QR = QP}}$ ${\color{Gray}{\left(\text{given}\right)}}$
3. ${\color{Gray}{\angle TQR = \angle SQP}}$ ${\color{Gray}{\left(\text{vert opp } \angle \text{s equal}\right)}}$

$\color{ForestGreen}{\therefore \Delta TQR \equiv \Delta SQP}$ $\color{ForestGreen}{\left(\text{SAS}\right)}$.

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. Which of these cases could you prove from the information given?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were given that $JN = MN$, so we have a pair of equal sides.

We were also told that $JK \parallel LM$, so we can use alternate angles between parallel lines to prove that $\hat{J} = \hat{M}$. So we have a pair of equal angles.

We can use vertically opposite angles to prove that $J\hat{N}L = M\hat{N}L$. So we have another pair of equal angles.

This matches the SAA (side, angle, angle) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the following format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles. The working to prove each of these three sides or angles is labelled 1, 2, 3.
• Conclude, and give a reason for the congruency.

In $\color{red}{\Delta JNK}$ and $\color{red}{\Delta MNL}$:

1. $\color{Gray}{JN = MN}$ $\color{Gray}{\left(\text{given}\right)}$
2. ${\color{Gray}{\angle J = \angle M}}$ ${\color{Gray}{\left(\text{alt } \angle \text{s; } JK \parallel LM\right)}}$
3. ${\color{Gray}{\angle JNK = \angle MNL}}$ ${\color{Gray}{\left(\text{vert opp } \angle \text{s equal}\right)}}$

$\color{ForestGreen}{\therefore \Delta JNK \equiv \Delta MNL}$ $\color{ForestGreen}{\left(\text{SAA}\right)}$.

Congruent triangles are exactly the same in size and shape. The order in which you write down the vertices of the second triangle must follow the order of the matching vertices in the first triangle. It might not be in alphabetical order!

You have already been told that the two triangles are congruent. All you need to do is decide what order the vertices of the second triangle should be written in. The equal sides and angles tell you which vertices match up between the two triangles. Write the matching vertices in the same place in the triangle's name.

The first triangle was named for us in the order $ABC$. We can see that the angle $\hat{B}$ is equal to angle $\hat{F}$. Therefore $B$ and $F$ must be in the same position in the triangle's names.

We can also see that the side $AB$ is equal in length to the side $DF$. Therefore, $A$ must be in the same position as $D$.

There is only one vertex $(E)$ left in our second triangle: it will come at the end of the triangle's name.

We can see that we have named the second triangle in such a way that the equal sides come in the same order for both triangles.

Two of the sides were given to be equal, as well as the angle in between the given sides. We use the reason 'side, angle, side', which is abbreviated to 'SAS'.

Congruent triangles are exactly the same in size and shape. The order in which you write down the vertices of the second triangle must follow the order of the matching vertices in the first triangle. It might not be in alphabetical order!

You have already been told that the two triangles are congruent. All you need to do is decide what order the vertices of the second triangle should be written in. The equal sides and angles tell you which vertices match up between the two triangles. Write the matching vertices in the same place in the triangle's name.

The first triangle was named for us in the order $ABC$. We can see that angle $\hat{C}$ is equal to angle $\hat{F}$, because they are both $90°$. Therefore $C$ and $F$ must be in the position in the triangles' names.

We can also see that side $BC$ is equal in length to $DF$. Therefore, we must place the letter $D$ in the same position as $B$.

There is only one vertex $(E)$ left in our second triangle: it will fill in the last empty space.

We can see that we have named the triangle in such a way that the equal angles and equal sides come in the same order for both triangles.

A $90°$ angle was given in both triangles, together with the hypotenuse and one other side. We use the reason '$90°$, hypotenuse, side', which is abbreviated to '90°HS'.

Congruent triangles are exactly the same in size and shape. The order in which you write down the vertices of the second triangle must follow the order of the matching vertices in the first triangle. It might not be in alphabetical order!

You have already been told that the two triangles are congruent. All you need to do is decide what order the vertices of the second triangle should be written in. The equal sides and angles tell you which vertices match up between the two triangles. Write the matching vertices in the same place in the triangle's name.

The first triangle was named for us in the order $ABC$. We can see that angle $\hat{B}$ is equal to angle $\hat{Y}$. Therefore, $B$ and $Y$ must be in the same position in the triangles' names. In the same way, angle $\hat{C}$ is equal to $\hat{Z}$. Therefore $C$ and $Z$ must be in the same position in the triangles' names.

There is only one vertex $(X)$ left in our second triangle: it will fill in the last empty space.

We can see that we have named the second triangle in such a way that the equal angles and equal sides come in the same order for both triangles.

One side and two angles were given to be equal, so we use the reason 'side, angle, angle'. This is abbreviated to 'SAA'.

To decide on the correct proof, you first need to decide what the proof should look like. In order to prove congruency, you need pairs of equal sides or angles. What geometry fact can you use to determine the third angle in one of the triangles?

We are looking for a proof that correctly demonstrates one of the four cases for congruency (SSS, SAA, SAS, $90°$HS).

✘ Option A

In Step 2, this options says that the reason why $K\hat{M}J = L\hat{K}M = 54°$ is because the two angles are alternate on parallel lines. But, we have not been told that $KL \parallel JM$, so we cannot use this fact (even if it looks like the lines could be parallel). If we want to use it, we have to prove it first. We can only use the information that we have been given, or that we have proven using our geometry facts.

✘ Option B

In Step 2, this option says that the reason why $\hat{J} = \hat{L}$ is because the two triangles are congruent. Since we haven't proved that the triangles are congruent yet, we cannot use the fact that they are congruent. In Geometry we have to know something for certain before we can use it to prove something else.

✔ Option C

This option correctly demonstrates one of the four cases for congruency. It uses the given information and sum of angles in a triangle to prove one pair of matching sides and two pairs of matching angles (SAA).

✘ Option D

This option correctly proves that all three angles in the two triangles match up. But, this is a case for similarity: it does not prove congruency. It is possible for two triangles to have the same angles, while being different sizes.

To decide on the correct proof, you first need to decide what the proof should look like. In order to prove congruency, you need pairs of equal sides or angles. What geometry fact can you use to determine the third angle in one of the triangles?

We are looking for a proof that correctly demonstrates one of the four cases for congruency (SSS, SAA, SAS, $90°$HS).

✘ Option A

This option correctly proves that all three angles in the two triangles match up. But, this is a case for similarity: it does not prove congruency. It is possible for two triangles to have the same angles, while being different sizes.

✘ Option B

In Step 2, this options says that the reason why $A\hat{D}B = C\hat{A}D = 41°$ is because the two angles are alternate on parallel lines. But, we have not been told that $AC \parallel BD$, so we cannot use this fact (even if it looks like the lines could be parallel). If we want to use it, we have to prove it first. We can only use the information that we have been given, or that we have proven using our geometry facts.

✘ Option C

In Step 2, this option says that the reason why $\hat{B} = \hat{C}$ is because the two triangles are congruent. Since we haven't proved that the triangles are congruent yet, we cannot use the fact that they are congruent. In Geometry we have to know something for certain before we can use it to prove something else.

✔ Option D

This option correctly demonstrates one of the four cases for congruency. It uses the given information and sum of angles in a triangle to prove one pair of matching sides and two pairs of matching angles (SAA).

To decide on the correct proof, you first need to decide what the proof should look like. In order to prove congruency, you need pairs of equal sides or angles. What geometry fact can you use to determine the third side in one of the right-angled triangles?

We are looking for a proof that correctly demonstrates one of the four cases for congruency (SSS, SAA, SAS, $90°$HS).

✘ Option A

In Step 2, this option says that $\hat{P}$ is common. But, a common angle is an angle that is shared by two triangles. In this case, there are two different angles at Point $P$: $R\hat{P}S$ and $Q\hat{P}S$. They might be equal, or they might not. They are definitely not common, as each triangle has its own angle at $P$.

✘ Option B

This proof does not include a reason for the congruency on the last line, so it cannot be the correct choice.

✘ Option C

In Step 3, this option says that the reason why $PQ = 10$ is because the two triangles are congruent. But, this is what we are trying to prove. We cannot use it as a fact to prove itself.

✔ Option D

This option correctly demonstrates one of the four cases for congruency. It uses the given information and the theorem of Pythagoras to prove one pair of matching $90°$ angles, one pair of hypotenuses, and one pair of matching sides (${\color{orange}{90°}}$HS).

There are four cases for congruency: SSS, SAS, SAA, and 90°HS. Which of the options given to you does not satisfy any of these cases, when you compare it to $\Delta DEF$?

Although the triangles all look like they might be congruent to $\Delta DEF$, we only know for sure that they are congruent if they satisfy one of the cases for congruency.

Triangle 1

Two of the sides of this triangle match up to sides of $\Delta DEF$, and the angles in between these two sides also match up. Therefore, the two triangles are congruent, because of SAS.

Triangle 2

Each side of this triangle matches up with one of the sides of $\Delta DEF$. Therefore, the two triangles are congruent, because of SSS.

Triangle 3

One of the sides of this triangle matches up with a side of $\Delta DEF$. Two of the angles of this triangle match up with angles in $\Delta DEF$. Therefore, the two triangles are congruent, because of SAA.

Triangle 4

Although this triangle has one side and one angle which match up to a side and angle in $\Delta DEF$, this is not enough to know for sure that the triangles are congruent. We need at least one more side or angle pair.

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Triangle 4

There are four cases for congruency: SSS, SAS, SAA, and 90°HS. Which of the options given to you does not satisfy any of these cases, when you compare it to $\Delta ABC$?

Although the triangles all look like they might be congruent to $\Delta ABC$, we only know for sure that they are congruent if they satisfy one of the cases for congruency.

Triangle 1

Two of the sides of this triangle match up to sides of $\Delta ABC$, and the angles in between these two sides also match up. Therefore, the two triangles are congruent, because of SAS.

Triangle 2

One of the sides of this triangle matches up with a side of $\Delta ABC$. Two of the angles of this triangle match up with angles in $\Delta ABC$. Therefore, the two triangles are congruent, because of SAA.

Triangle 4

Each side of this triangle matches up with one of the sides of $\Delta ABC$. Therefore, the two triangles are congruent, because of SSS.

Triangle 3

Although all three of the angles in this triangles are equal to the angles in $\Delta ABC$, the two triangles could still have completely different sizes.

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Triangle 3

There are four cases for congruency: SSS, SAS, SAA, and 90°HS. Which of the options given to you does not satisfy any of these cases, when you compare it to $\Delta ABC$?

Although the triangles all look like they might be congruent to $\Delta ABC$, we only know for sure that they are congruent if they satisfy one of the cases for congruency.

Triangle 2

Each side of this triangle matches up with one of the sides of $\Delta ABC$. Therefore, the two triangles are congruent, because of SSS.

Triangle 3

One of the sides of this triangle matches up with a side of $\Delta ABC$. Two of the angles of this triangle match up with angles in $\Delta ABC$. Therefore, the two triangles are congruent, because of SAA.

Triangle 4

Two of the sides of this triangle match up to sides of $\Delta ABC$, and the angles in between these two sides also match up. Therefore, the two triangles are congruent, because of SAS.

Triangle 1

Although all three of the angles in this triangles are equal to the angles in $\Delta ABC$, the two triangles could still have completely different sizes.

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Triangle 1

The distance from the centre of a circle to its circumference is always the same length. It is called the radius. Which of the four congruency cases could you prove using this fact and the other given information?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $OC = OB$, so we have a pair of equal sides.

We were also told that $EB$ and $CD$ are straight lines, so they form vertically opposite angles where they intersect. We can use this fact to prove that $E\hat{O}C = D\hat{O}B$ . So, we have a pair of equal angles.

Finally, we were told that $O$ is the centre of the circle. The distance from the centre of a circle to its circumference is always the same length. It is called the radius. We can use this fact to prove that $OE = OD$. So, we have another pair of equal sides.

The angle is between the two sides, so this matches the SAS (side, angle, side) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the following format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles.
• Conclude, and give a reason for the congruency. Make sure that you label the triangles in the matching order.

In $\color{red}{\Delta OEC}$ and $\color{red}{\Delta ODB}$:

1. $\color{Gray}{OC = OB}$ $\color{Gray}{\left(\text{given}\right)}$
2. $\color{Gray}{E\hat{O}C = D\hat{O}B}$ $\color{Gray}{\left(\text{vert opp } \angle \text{s equal}\right)}$
3. $\color{Gray}{OE = OD}$ $\color{Gray}{\left(\text{radii}\right)}$

$\color{ForestGreen}{\therefore \Delta OEC \equiv \Delta ODB}$ $\color{ForestGreen}{\left(\text{SAS}\right)}$.

The distance from the centre of a circle to its circumference is always the same length. It is called the radius. Which of the four congruency cases could you prove using this fact and the other given information?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $KM = JM$, so we have a pair of equal sides.

We were also told that $O$ is the centre of the circle. The distance from the centre of a circle to its circumference is always the same length. It is called the radius. We can use this fact to prove that $OK = OJ$, so we have a pair of equal sides.

We can see that the line $OM$ is a side of both triangles. When the same side is shared by two triangles, we say that the side is common. So, we have a pair of equal sides.

This matches the SSS (side, side, side) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the following format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles.
• Conclude, and give a reason for the congruency. Make sure that you label the triangles in the matching order.

In $\color{red}{\Delta OKM}$ and $\color{red}{\Delta OJM}$:

1. $\color{Gray}{KM = JM}$ $\color{Gray}{\left(\text{given}\right)}$
2. $\color{Gray}{OK = OJ}$ $\color{Gray}{\left(\text{radii}\right)}$
3. $\color{Gray}{OM \text{ is common}}$

$\color{ForestGreen}{\therefore \Delta OKM \equiv \Delta OJM}$ $\color{ForestGreen}{\left(\text{SSS}\right)}$.

The distance from the centre of a circle to its circumference is always the same length. It is called the radius. Which of the four congruency cases could you prove using this fact and the other given information?

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $OR \; \bot \; PQ$, so we know that $O\hat{R}P = O\hat{R}Q = 90°$. So we have a pair of $\color{teal}{90°}$ angles.

We were also told that $O$ is the centre of the circle. The distance from the centre of a circle to its circumference is always the same length. It is called the radius. We can use this fact to prove that $OP = OQ$, so we have a pair of equal sides. However, $OP$ is the hypotenuse of $\Delta OPR$, and $OQ$ is the hypotenuse of $\Delta OQR$, because they are opposite the $90°$ angles. So we have a pair of equal hypotenuses.

We can see that the line $OR$ is a side of both triangles. When the same side is shared by two triangles, we say that the side is common. So, we have a pair of equal sides.

This matches the 90$\color{teal}{°}$HS ($90°$, hypotenuse, side) case for congruency.

Now that we have identified the congruency case that can be proven, we need to write out the formal proof. Congruency proofs must always have the following format:

• State the triangles in which you will prove congruency.
• Prove the congruency by showing three pairs of matching sides or angles.
• Conclude, and give a reason for the congruency. Make sure that you label the triangles in the matching order.

In $\color{red}{\Delta OPR}$ and $\color{red}{\Delta OQR}$:

1. $\color{Gray}{O\hat{R}P = O\hat{R}Q = 90°}$ $\color{Gray}{\left(\text{given}\right)}$
2. $\color{Gray}{OP = OQ}$ $\color{Gray}{\left(\text{radii}\right)}$
3. $\color{Gray}{OR \text{ is common}}$

$\color{ForestGreen}{\therefore \Delta OPR \equiv \Delta OQR}$ $\color{ForestGreen}{\left(90°\text{HS}\right)}$.

If two triangles are congruent, they are exactly the same size and shape. What conclusions can you draw about the lengths of sides and sizes of angles in the two triangles?

If two triangles are congruent, their matching angles and sides will be equal.

We were told that $\Delta KLJ \equiv \Delta ABC$, which means that ${\color{red}{\hat{K}}}$ is the same size as ${\color{red}{\hat{A}}}$. In the same way, ${\color{blue}{\hat{L}}}$ is the same size as ${\color{blue}{\hat{B}}}$, and ${\color{ForestGreen}{\hat{J}}}$ is the same size as ${\color{ForestGreen}{\hat{C}}}$.

This means that $x = 54°$, because $K$ matches $A$. We use the reason $\left(\Delta KLJ \equiv \Delta ABC\right)$ to remind ourselves that the angles are only equal because the triangles are congruent.

In the same way, ${\color{red}{KL}}$ is the same length as ${\color{red}{AB}}$, ${\color{blue}{LJ}}$ is the same length as ${\color{blue}{BC}}$, and ${\color{ForestGreen}{KJ}}$ is the same length as ${\color{ForestGreen}{AC}}$.

This means that $y = \text{9,76}\ \text{units}$, because $JK$ is equivalent to $CA$. We use the reason $\left(\Delta KLJ \equiv \Delta ABC\right)$ to remind ourselves that the sides are only equal because the triangles are congruent.

If two triangles are congruent, they are exactly the same size and shape. What conclusions can you draw about the lengths of sides and sizes of angles in the two triangles?

If two triangles are congruent, their matching angles and sides will be equal.

We were told that $\Delta YZX \equiv \Delta JKL$, which means that ${\color{red}{\hat{Y}}}$ is the same size as ${\color{red}{\hat{J}}}$. In the same way, ${\color{blue}{\hat{Z}}}$ is the same size as ${\color{blue}{\hat{K}}}$, and ${\color{ForestGreen}{\hat{X}}}$ is the same size as ${\color{ForestGreen}{\hat{L}}}$.

This means that $a = 52°$, because $X$ matches $L$. We use the reason $\left(\Delta YZX \equiv \Delta JKL\right)$ to remind ourselves that the angles are only equal because the triangles are congruent.

In the same way, ${\color{red}{YZ}}$ is the same length as ${\color{red}{JK}}$, ${\color{blue}{ZX}}$ is the same length as ${\color{blue}{KL}}$, and ${\color{ForestGreen}{YX}}$ is the same length as ${\color{ForestGreen}{JL}}$.

This means that $b = \text{17,35}\ \text{m}$, because $XY$ is equivalent to $LJ$. We use the reason $\left(\Delta YZX \equiv \Delta JKL\right)$ to remind ourselves that the sides are only equal because the triangles are congruent.

If two triangles are congruent, they are exactly the same size and shape. What conclusions can you draw about the lengths of sides and sizes of angles in the two triangles?

If two triangles are congruent, their matching angles and sides will be equal.

We were told that $\Delta PQR \equiv \Delta JKL$, which means that ${\color{red}{\hat{P}}}$ is the same size as ${\color{red}{\hat{J}}}$. In the same way, ${\color{blue}{\hat{Q}}}$ is the same size as ${\color{blue}{\hat{K}}}$, and ${\color{ForestGreen}{\hat{R}}}$ is the same size as ${\color{ForestGreen}{\hat{L}}}$.

This means that $a = 42°$, because $R$ matches $L$. We use the reason $\left(\Delta PQR \equiv \Delta JKL\right)$ to remind ourselves that the angles are only equal because the triangles are congruent.

In the same way, ${\color{red}{PQ}}$ is the same length as ${\color{red}{JK}}$, ${\color{blue}{QR}}$ is the same length as ${\color{blue}{KL}}$, and ${\color{ForestGreen}{PR}}$ is the same length as ${\color{ForestGreen}{JL}}$.

This means that $b = \text{8,49}\ \text{m}$, because $PQ$ is equivalent to $JK$. We use the reason $\left(\Delta PQR \equiv \Delta JKL\right)$ to remind ourselves that the sides are only equal because the triangles are congruent.

You cannot determine the size of $\hat{D}$ directly using the geometry facts that you know. The only option is to prove that the two triangles are congruent, and then match up the angles.

We cannot determine the size of $\hat{D}$ directly using the geometry facts that we know. First we must prove that the two triangles are congruent, and then use this fact to match up the angles.

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $OC = OD$, so we have a pair of equal sides.

We were also told that $ED$ and $CB$ are straight lines, so they form vertically opposite angles where they intersect. We can use this fact to prove that $E\hat{O}C = B\hat{O}D$ . So, we have a pair of equal angles.

Finally, we were told that $O$ is the centre of the circle. The distance from the centre of a circle to its circumference is always the same length. It is called the radius. We can use this fact to prove that $OE = OB$. So, we have another pair of equal sides.

The angle is between the two sides, so this matches the SAS (side, angle, side) case for congruency.

Now we can write the proof of the congruency:

In $\Delta OEC$ and $\Delta OBD$:

1. $OC = OD$ $\left(\text{given}\right)$
2. $E\hat{O}C = B\hat{O}D$ $\left(\text{vert opp } \angle \text{s equal}\right)$
3. ${\color{red}{OE = OB}}$ ${\color{red}{\left(\text{radii}\right)}}$

$\therefore \Delta OEC \equiv \Delta OBD$ $\left(\text{SAS}\right)$

The triangles are congruent, so their matching sides and angles are congruent.

You cannot determine the length of $BE$ directly using the geometry facts that you know. The only option is to prove that the two triangles are congruent, and then match up the sides.

We cannot determine the length of $BE$ directly using the geometry facts that we know. First we must prove that the two triangles are congruent, and then use this fact to match up the sides.

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $OE \; \bot \; CB$, so we know that $O\hat{E}C = O\hat{E}B = 90°$. So we have a pair of $\color{teal}{90°}$ angles.

We were also told that $O$ is the centre of the circle. The distance from the centre of a circle to its circumference is always the same length. It is called the radius. We can use this fact to prove that $OC = OB$, so we have a pair of equal sides. However, $OC$ is the hypotenuse of $\Delta OCE$, and $OB$ is the hypotenuse of $\Delta OBE$, because they are opposite the $90°$ angles. So we have a pair of equal hypotenuses.

We can see that the line $OE$ is a side of both triangles. When the same side is shared by two triangles, we say that the side is common. So, we have a pair of equal sides.

This matches the 90$\color{teal}{°}$HS ($90°$, hypotenuse, side) case for congruency.

Now we can write the proof of the congruency:

In $\Delta OCE$ and $\Delta OBE$:

1. $O\hat{E}C = O\hat{E}B = 90°$ $\left(\text{given}\right)$
2. $OC = OB$ ${\color{red}{\left(\text{radii}\right)}}$
3. $OE \text{ is common}$

$\therefore \Delta OCE \equiv \Delta OBE$ ${\color{red}{\left(90°\text{HS}\right)}}$

The triangles are congruent, so their matching sides and angles are congruent.

You cannot determine the size of $E\hat{O}B$ directly using the geometry facts that you know. The only option is to prove that the two triangles are congruent, and then match up the angles.

We cannot determine the size of $E\hat{O}B$ directly using the geometry facts that we know. First we must prove that the two triangles are congruent, and then use this fact to match up the angles.

The four cases for congruency are SSS, SAA, SAS, and $90°$HS. When we prove congruency, we must first decide which of the four cases we can prove before writing out the proof.

We were told that $EB = DB$, so we have a pair of equal sides.

We were also told that $O$ is the centre of the circle. The distance from the centre of a circle to its circumference is always the same length. It is called the radius. We can use this fact to prove that $OE = OD$, so we have a pair of equal sides.

We can see that the line $OB$ is a side of both triangles. When the same side is shared by two triangles, we say that the side is common. So, we have a pair of equal sides.

This matches the SSS (side, side, side) case for congruency.

Now we can write the proof of the congruency:

In $\Delta OEB$ and $\Delta ODB$:

1. $EB = DB$ $\left(\text{given}\right)$
2. $OE = OD$ $\left(\text{radii}\right)$
3. ${\color{red}{OB \text{ is common}}}$

$\therefore \Delta OEB \equiv \Delta ODB$ ${\color{red}{\left(\text{SSS}\right)}}$

The triangles are congruent, so their matching sides and angles are congruent.

Since the triangles are congruent, their matching sides will be equal. Can you use a geometry reason to find the length of the side which matches $q$?

If two triangles are congruent, their matching angles and sides will be equal.

We were told that $\Delta JKL \equiv \Delta XZY$ which means that ${\color{red}{JK}}$ is the same length as ${\color{red}{XZ}}$. In the same way, ${\color{blue}{KL}}$ is the same length as ${\color{blue}{ZY}}$, and ${\color{ForestGreen}{JL}}$ is the same length as ${\color{ForestGreen}{XY}}$. The matching angles are also equal, but for this question we only need to think about the sides.

This means that $q = KL$, because $ZY$ is equivalent to $KL$. We use the reason $\left(\Delta JKL \equiv \Delta XZY\right)$ to remind ourselves that the sides are only equal because the triangles are congruent.

We don't have enough information to find the size of $q$ in triangle $XZY$.

But we know that $q = KL$, and we have enough information to find $KL$ in triangle $JKL$, using the theorem of Pythagoras.

Since the triangles are congruent, their matching angles will be equal. Can you use a geometry reason to find the size of the angle which matches $k$?

If two triangles are congruent, their matching angles and sides will be equal.

We were told that $\Delta ACB \equiv \Delta QRP$ which means that ${\color{red}{\hat{A}}}$ is the same size as ${\color{red}{\hat{Q}}}$. In the same way, ${\color{blue}{\hat{C}}}$ is the same size as ${\color{blue}{\hat{R}}}$, and ${\color{ForestGreen}{\hat{B}}}$ is the same size as ${\color{ForestGreen}{\hat{P}}}$. The matching sides are also equal, but for this question we only need to think about the angles.

This means that $k = \hat{A}$, because $Q$ matches $A$. We use the reason $\left(\Delta ACB \equiv \Delta QRP\right)$ to remind ourselves that the angles are only equal because the triangles are congruent.

We don't have enough information to find the size of $k$ in triangle $QRP$.

But we know that $k = \hat{A}$ and we have enough information to find $\hat{A}$ in triangle $ACB$, using sum of angles in a triangle.

Since the triangles are congruent, their matching angles will be equal. Can you use a geometry reason to find the size of the angle which matches $z$?

If two triangles are congruent, their matching angles and sides will be equal.

We were told that $\Delta BAC \equiv \Delta KLJ$ which means that ${\color{red}{\hat{B}}}$ is the same size as ${\color{red}{\hat{K}}}$. In the same way, ${\color{blue}{\hat{A}}}$ is the same size as ${\color{blue}{\hat{L}}}$, and ${\color{ForestGreen}{\hat{C}}}$ is the same size as ${\color{ForestGreen}{\hat{J}}}$. The matching sides are also equal, but for this question we only need to think about the angles.

This means that $z = \hat{B}$, because $K$ matches $B$. We use the reason $\left(\Delta BAC \equiv \Delta KLJ\right)$ to remind ourselves that the angles are only equal because the triangles are congruent.